Question

помогите решить
дам (30 б)​

Answer 1

(1/9)^x - 4*(1/3)^(x - 1) + 27 ≤ 0

(1/9)^x - 4*3/3*(1/3)^(x - 1) + 27 ≤ 0

(1/9)^x - 12*(1/3)^x + 27 ≤ 0

(1/3)^x = y > 0

y² - 12y + 27 ≤ 0

D = 12² - 4*27 = 36

y12 = (12 +- 6)/2 = 3   9

(y - 3)(y - 9)≤ 0

++++++++[3] ------------- [9] ++++++++

y∈ [3, 9]

y ≤ 9  

(1/3)^x ≤ 9

3^-x ≤ 3^2

-x ≤ 2

x ≥ -2    

y ≥ 3

(1/3)^x ≥ 3

3^-x ≥ 3

-x ≥ 1

x ≤ -1

ответ x ∈ [-2, -1]